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Chapter 12: Problem 16

Give an example of a Galois extension of \(\mathbb{Q}\) such that the Galois group is as indicated. \(\operatorname{Gal}(E / Q)=\mathbb{Z}_{4}\)

Short Answer

Expert verified
The extension \( \mathbb{Q}(\sqrt[4]{2}, i) \) is a Galois extension with Galois group \( \mathbb{Z}_4 \).

Step by step solution

01

Understand Galois Extensions

A Galois extension is a field extension that is both normal and separable. The Galois group of the extension is a group of field automorphisms that maps the field onto itself while fixing the base field.
02

Identify a Suitable Polynomial

To construct a Galois extension of \( \mathbb{Q} \) with a Galois group isomorphic to \( \mathbb{Z}_4 \), consider quartic polynomials. An easy candidate is \( x^4 - 2 \), as it has distinct roots, making it separable.
03

Check Normality

The polynomial \( x^4 - 2 \) is normal over \( \mathbb{Q} \) because it can be written as \( x^4 = 2 \), which has all roots expressible in terms of radicals. In particular, its roots include \( \sqrt[4]{2} \), \(-\sqrt[4]{2} \), \(i\sqrt[4]{2} \), and \(-i\sqrt[4]{2} \).
04

Determine the Galois Group

The extension \( E = \mathbb{Q}(\sqrt[4]{2}, i) \) is a splitting field of \( x^4 - 2 \). The degree of this extension is 4 because the minimal polynomial over \( \mathbb{Q} \) has degree 4. Hence, the Galois group is isomorphic to \( \mathbb{Z}_4 \), which matches the required Galios Group order of 4.
05

Verify Solution Correctness

Since \( E = \mathbb{Q}(\sqrt[4]{2}) \) has degree 4, and considering that the extension is both normal and separable, the Galois group \( \operatorname{Gal}(E/\mathbb{Q}) \) is indeed isomorphic to \( \mathbb{Z}_4 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Galois group
In the world of field extensions, the Galois group plays a crucial role. It's essentially a group of all field automorphisms, which are functions that map a field back onto itself while keeping the base field unchanged. Think of it as the symmetries of a field extension. Each automorphism in the Galois group provides a new perspective on the solutions within a given polynomial.
  • An automorphism is just like a symmetry in geometry, showing the various ways to view a field without altering its core features.
  • The size of the Galois group, also known as the order of the group, typically matches the degree of the field extension.
  • In our example, we aimed for a Galois group isomorphic to \(\mathbb{Z}_4\), indicating a cyclic group of order 4.
Understanding the Galois group is vital as it helps elucidate whether the field extensions are solvable by radicals, that is, if roots of polynomials can be expressed using a combination of arithmetic operations and root extractions.
field extension
A field extension occurs when you enlarge a smaller field by introducing new elements. These new elements allow you to "solve" more equations within your broader mathematical system.
  • Think of a field extension as expanding your mathematical toolkit.
  • For example, moving from the rational numbers \(\mathbb{Q}\) to the complex numbers \(\mathbb{C}\) via the imaginary unit \(i\) is an extension.
  • In the exercise, the extension \(E = \mathbb{Q}(\sqrt[4]{2}, i)\) is constructed by adding \(\sqrt[4]{2}\) and \(i\), thus allowing the solving of the polynomial \(x^4 - 2\).
Field extensions can be finite or infinite, but in our case, we deal predominantly with finite extensions, like our Galois extension of degree 4.
normal and separable
For a field extension to be called Galois, it must be both normal and separable. But what does this mean?
  • A normal extension is one where every polynomial that has a root in the extended field splits completely into linear factors.
  • This means that all the solutions (or roots) of the polynomial can be expressed within that extended field.
  • A separable extension, on the other hand, guarantees that the polynomial has distinct roots, without any repetitions.
In the exercise, the polynomial \(x^4 - 2\) is both normal and separable over \(\mathbb{Q}\):
  • The roots \(\sqrt[4]{2}\), \(-\sqrt[4]{2}\), \(i\sqrt[4]{2}\), and \(-i\sqrt[4]{2}\) confirm this situation.
  • These roots are distinct, ensuring separability.
  • The polynomial splits completely into linear factors because it can be expressed as \((x-\sqrt[4]{2})(x+\sqrt[4]{2})(x-i\sqrt[4]{2})(x+i\sqrt[4]{2})\).
quartic polynomials
Quartic polynomials are equations of degree 4, which means they are typically expressed as \(ax^4 + bx^3 + cx^2 + dx + e = 0\). These polynomials can often present interesting challenges and solutions in algebra.
  • The degree of a quartic polynomial indicates how many solutions or roots it may have, considering the right allowing for complex numbers.
  • Quartic polynomials are notable because of their placement in Galois theory, where solving them typically requires advanced techniques and field extensions.
  • For the exercise at hand, the quartic polynomial \(x^4 - 2\) ensures a rich structure that is both normal and separable, making it an excellent candidate for constructing a Galois extension.
The process involves finding all the roots and observing how they interact with \(\mathbb{Q}\). Each of these roots helps form a complete understanding of the underlying field structures, symmetries, and extensions.

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Most popular questions from this chapter

Let \(E\) be a Galois extension of a field \(F\) with Gal(E/F) cyclic. Show that for each divisor \(d\) of \([E: F]\) there exists exactly one intermediate field \(F \subseteq K \subseteq E\) such that \([K: F]=d\)

Show that \(Q(\sqrt{3})\) and \(Q(\sqrt{5})\) are not isomorphic fields.

Determine whether the indicated field extension is a Galois extension. $$ \mathrm{Q}\left(\sqrt[3]{2}, i \sin \frac{2 \pi}{3}\right) \text { over } \mathrm{Q} $$

Let \(\alpha\) be a positive real number that is constructible by straightedge and compass, and \(f(x)\) its minimal polynomial over \(Q\). Show that \(f(x)\) is solvable by radicals over \(Q\)

Let \(f(x) \in F[x]\) be an irreducible polynomial of degree \(n\) over a field \(F\), and let \(E\) be its splitting ficld. Show that (a) \(n\) divides \([E: F]\). (b) If \(f(x)\) is separable, then \(n\) divides \(|\operatorname{Gal}(E / F)|\)
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